YES(?,O(n^1)) * Step 1: TrivialSCCs WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. start(A,B,C,D) -> stop1(A,B,C,D) [A + -1*D >= 0 (?,1) && D >= 0 && C + D >= 0 && B + D >= 0 && A + D >= 0 && C >= 0 && B + C >= 0 && A + C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && D = 0] 1. start(A,B,C,D) -> cont1(A,B,C,D) [A + -1*D >= 0 (?,1) && D >= 0 && C + D >= 0 && B + D >= 0 && A + D >= 0 && C >= 0 && B + C >= 0 && A + C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && D >= 1 && A >= D] 2. cont1(A,B,C,D) -> stop2(A,B,1,-1 + D) [A + -1*D >= 0 (?,1) && -1 + D >= 0 && -1 + C + D >= 0 && -1 + B + D >= 0 && -2 + A + D >= 0 && C >= 0 && B + C >= 0 && -1 + A + C >= 0 && B >= 0 && -1 + A + B >= 0 && -1 + A >= 0 && D >= 1 && A >= D && C = 0] 3. cont1(A,B,C,D) -> a(A,B,-1 + C,D) [A + -1*D >= 0 (?,1) && -1 + D >= 0 && -1 + C + D >= 0 && -1 + B + D >= 0 && -2 + A + D >= 0 && C >= 0 && B + C >= 0 && -1 + A + C >= 0 && B >= 0 && -1 + A + B >= 0 && -1 + A >= 0 && C >= 1 && D >= 1 && A >= D] 4. a(A,B,C,D) -> b(A,B,E,-1 + D) [A + -1*D >= 0 (?,1) && -1 + D >= 0 && -1 + C + D >= 0 && -1 + B + D >= 0 && -2 + A + D >= 0 && C >= 0 && B + C >= 0 && -1 + A + C >= 0 && B >= 0 && -1 + A + B >= 0 && -1 + A >= 0 && A >= D && D >= 1] 5. b(A,B,C,D) -> start(A,B,C,D) [-1 + A + -1*D >= 0 (?,1) && D >= 0 && B + D >= 0 && -1 + A + D >= 0 && B >= 0 && -1 + A + B >= 0 && -1 + A >= 0 && C >= 0 && A >= 1 + D] 6. b(A,B,C,D) -> stop3(A,B,C,D) [-1 + A + -1*D >= 0 (?,1) && D >= 0 && B + D >= 0 && -1 + A + D >= 0 && B >= 0 && -1 + A + B >= 0 && -1 + A >= 0 && 0 >= 1 + C && A >= 1 + D] 7. start0(A,B,C,D) -> start(A,B,B,A) [A >= 0 && B >= 0] (1,1) Signature: {(a,4);(b,4);(cont1,4);(start,4);(start0,4);(stop1,4);(stop2,4);(stop3,4)} Flow Graph: [0->{},1->{2,3},2->{},3->{4},4->{5,6},5->{0,1},6->{},7->{0,1}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 2: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. start(A,B,C,D) -> stop1(A,B,C,D) [A + -1*D >= 0 (1,1) && D >= 0 && C + D >= 0 && B + D >= 0 && A + D >= 0 && C >= 0 && B + C >= 0 && A + C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && D = 0] 1. start(A,B,C,D) -> cont1(A,B,C,D) [A + -1*D >= 0 (?,1) && D >= 0 && C + D >= 0 && B + D >= 0 && A + D >= 0 && C >= 0 && B + C >= 0 && A + C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && D >= 1 && A >= D] 2. cont1(A,B,C,D) -> stop2(A,B,1,-1 + D) [A + -1*D >= 0 (1,1) && -1 + D >= 0 && -1 + C + D >= 0 && -1 + B + D >= 0 && -2 + A + D >= 0 && C >= 0 && B + C >= 0 && -1 + A + C >= 0 && B >= 0 && -1 + A + B >= 0 && -1 + A >= 0 && D >= 1 && A >= D && C = 0] 3. cont1(A,B,C,D) -> a(A,B,-1 + C,D) [A + -1*D >= 0 (?,1) && -1 + D >= 0 && -1 + C + D >= 0 && -1 + B + D >= 0 && -2 + A + D >= 0 && C >= 0 && B + C >= 0 && -1 + A + C >= 0 && B >= 0 && -1 + A + B >= 0 && -1 + A >= 0 && C >= 1 && D >= 1 && A >= D] 4. a(A,B,C,D) -> b(A,B,E,-1 + D) [A + -1*D >= 0 (?,1) && -1 + D >= 0 && -1 + C + D >= 0 && -1 + B + D >= 0 && -2 + A + D >= 0 && C >= 0 && B + C >= 0 && -1 + A + C >= 0 && B >= 0 && -1 + A + B >= 0 && -1 + A >= 0 && A >= D && D >= 1] 5. b(A,B,C,D) -> start(A,B,C,D) [-1 + A + -1*D >= 0 (?,1) && D >= 0 && B + D >= 0 && -1 + A + D >= 0 && B >= 0 && -1 + A + B >= 0 && -1 + A >= 0 && C >= 0 && A >= 1 + D] 6. b(A,B,C,D) -> stop3(A,B,C,D) [-1 + A + -1*D >= 0 (1,1) && D >= 0 && B + D >= 0 && -1 + A + D >= 0 && B >= 0 && -1 + A + B >= 0 && -1 + A >= 0 && 0 >= 1 + C && A >= 1 + D] 7. start0(A,B,C,D) -> start(A,B,B,A) [A >= 0 && B >= 0] (1,1) Signature: {(a,4);(b,4);(cont1,4);(start,4);(start0,4);(stop1,4);(stop2,4);(stop3,4)} Flow Graph: [0->{},1->{2,3},2->{},3->{4},4->{5,6},5->{0,1},6->{},7->{0,1}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(a) = x4 p(b) = 1 + x4 p(cont1) = x4 p(start) = x4 p(start0) = x1 p(stop1) = x4 p(stop2) = x3 + x4 p(stop3) = 1 + x4 Following rules are strictly oriented: [-1 + A + -1*D >= 0 ==> && D >= 0 && B + D >= 0 && -1 + A + D >= 0 && B >= 0 && -1 + A + B >= 0 && -1 + A >= 0 && C >= 0 && A >= 1 + D] b(A,B,C,D) = 1 + D > D = start(A,B,C,D) Following rules are weakly oriented: [A + -1*D >= 0 ==> && D >= 0 && C + D >= 0 && B + D >= 0 && A + D >= 0 && C >= 0 && B + C >= 0 && A + C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && D = 0] start(A,B,C,D) = D >= D = stop1(A,B,C,D) [A + -1*D >= 0 ==> && D >= 0 && C + D >= 0 && B + D >= 0 && A + D >= 0 && C >= 0 && B + C >= 0 && A + C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && D >= 1 && A >= D] start(A,B,C,D) = D >= D = cont1(A,B,C,D) [A + -1*D >= 0 ==> && -1 + D >= 0 && -1 + C + D >= 0 && -1 + B + D >= 0 && -2 + A + D >= 0 && C >= 0 && B + C >= 0 && -1 + A + C >= 0 && B >= 0 && -1 + A + B >= 0 && -1 + A >= 0 && D >= 1 && A >= D && C = 0] cont1(A,B,C,D) = D >= D = stop2(A,B,1,-1 + D) [A + -1*D >= 0 ==> && -1 + D >= 0 && -1 + C + D >= 0 && -1 + B + D >= 0 && -2 + A + D >= 0 && C >= 0 && B + C >= 0 && -1 + A + C >= 0 && B >= 0 && -1 + A + B >= 0 && -1 + A >= 0 && C >= 1 && D >= 1 && A >= D] cont1(A,B,C,D) = D >= D = a(A,B,-1 + C,D) [A + -1*D >= 0 ==> && -1 + D >= 0 && -1 + C + D >= 0 && -1 + B + D >= 0 && -2 + A + D >= 0 && C >= 0 && B + C >= 0 && -1 + A + C >= 0 && B >= 0 && -1 + A + B >= 0 && -1 + A >= 0 && A >= D && D >= 1] a(A,B,C,D) = D >= D = b(A,B,E,-1 + D) [-1 + A + -1*D >= 0 ==> && D >= 0 && B + D >= 0 && -1 + A + D >= 0 && B >= 0 && -1 + A + B >= 0 && -1 + A >= 0 && 0 >= 1 + C && A >= 1 + D] b(A,B,C,D) = 1 + D >= 1 + D = stop3(A,B,C,D) [A >= 0 && B >= 0] ==> start0(A,B,C,D) = A >= A = start(A,B,B,A) * Step 3: KnowledgePropagation WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. start(A,B,C,D) -> stop1(A,B,C,D) [A + -1*D >= 0 (1,1) && D >= 0 && C + D >= 0 && B + D >= 0 && A + D >= 0 && C >= 0 && B + C >= 0 && A + C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && D = 0] 1. start(A,B,C,D) -> cont1(A,B,C,D) [A + -1*D >= 0 (?,1) && D >= 0 && C + D >= 0 && B + D >= 0 && A + D >= 0 && C >= 0 && B + C >= 0 && A + C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && D >= 1 && A >= D] 2. cont1(A,B,C,D) -> stop2(A,B,1,-1 + D) [A + -1*D >= 0 (1,1) && -1 + D >= 0 && -1 + C + D >= 0 && -1 + B + D >= 0 && -2 + A + D >= 0 && C >= 0 && B + C >= 0 && -1 + A + C >= 0 && B >= 0 && -1 + A + B >= 0 && -1 + A >= 0 && D >= 1 && A >= D && C = 0] 3. cont1(A,B,C,D) -> a(A,B,-1 + C,D) [A + -1*D >= 0 (?,1) && -1 + D >= 0 && -1 + C + D >= 0 && -1 + B + D >= 0 && -2 + A + D >= 0 && C >= 0 && B + C >= 0 && -1 + A + C >= 0 && B >= 0 && -1 + A + B >= 0 && -1 + A >= 0 && C >= 1 && D >= 1 && A >= D] 4. a(A,B,C,D) -> b(A,B,E,-1 + D) [A + -1*D >= 0 (?,1) && -1 + D >= 0 && -1 + C + D >= 0 && -1 + B + D >= 0 && -2 + A + D >= 0 && C >= 0 && B + C >= 0 && -1 + A + C >= 0 && B >= 0 && -1 + A + B >= 0 && -1 + A >= 0 && A >= D && D >= 1] 5. b(A,B,C,D) -> start(A,B,C,D) [-1 + A + -1*D >= 0 (A,1) && D >= 0 && B + D >= 0 && -1 + A + D >= 0 && B >= 0 && -1 + A + B >= 0 && -1 + A >= 0 && C >= 0 && A >= 1 + D] 6. b(A,B,C,D) -> stop3(A,B,C,D) [-1 + A + -1*D >= 0 (1,1) && D >= 0 && B + D >= 0 && -1 + A + D >= 0 && B >= 0 && -1 + A + B >= 0 && -1 + A >= 0 && 0 >= 1 + C && A >= 1 + D] 7. start0(A,B,C,D) -> start(A,B,B,A) [A >= 0 && B >= 0] (1,1) Signature: {(a,4);(b,4);(cont1,4);(start,4);(start0,4);(stop1,4);(stop2,4);(stop3,4)} Flow Graph: [0->{},1->{2,3},2->{},3->{4},4->{5,6},5->{0,1},6->{},7->{0,1}] + Applied Processor: KnowledgePropagation + Details: We propagate bounds from predecessors. * Step 4: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. start(A,B,C,D) -> stop1(A,B,C,D) [A + -1*D >= 0 (1,1) && D >= 0 && C + D >= 0 && B + D >= 0 && A + D >= 0 && C >= 0 && B + C >= 0 && A + C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && D = 0] 1. start(A,B,C,D) -> cont1(A,B,C,D) [A + -1*D >= 0 (1 + A,1) && D >= 0 && C + D >= 0 && B + D >= 0 && A + D >= 0 && C >= 0 && B + C >= 0 && A + C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && D >= 1 && A >= D] 2. cont1(A,B,C,D) -> stop2(A,B,1,-1 + D) [A + -1*D >= 0 (1,1) && -1 + D >= 0 && -1 + C + D >= 0 && -1 + B + D >= 0 && -2 + A + D >= 0 && C >= 0 && B + C >= 0 && -1 + A + C >= 0 && B >= 0 && -1 + A + B >= 0 && -1 + A >= 0 && D >= 1 && A >= D && C = 0] 3. cont1(A,B,C,D) -> a(A,B,-1 + C,D) [A + -1*D >= 0 (1 + A,1) && -1 + D >= 0 && -1 + C + D >= 0 && -1 + B + D >= 0 && -2 + A + D >= 0 && C >= 0 && B + C >= 0 && -1 + A + C >= 0 && B >= 0 && -1 + A + B >= 0 && -1 + A >= 0 && C >= 1 && D >= 1 && A >= D] 4. a(A,B,C,D) -> b(A,B,E,-1 + D) [A + -1*D >= 0 (1 + A,1) && -1 + D >= 0 && -1 + C + D >= 0 && -1 + B + D >= 0 && -2 + A + D >= 0 && C >= 0 && B + C >= 0 && -1 + A + C >= 0 && B >= 0 && -1 + A + B >= 0 && -1 + A >= 0 && A >= D && D >= 1] 5. b(A,B,C,D) -> start(A,B,C,D) [-1 + A + -1*D >= 0 (A,1) && D >= 0 && B + D >= 0 && -1 + A + D >= 0 && B >= 0 && -1 + A + B >= 0 && -1 + A >= 0 && C >= 0 && A >= 1 + D] 6. b(A,B,C,D) -> stop3(A,B,C,D) [-1 + A + -1*D >= 0 (1,1) && D >= 0 && B + D >= 0 && -1 + A + D >= 0 && B >= 0 && -1 + A + B >= 0 && -1 + A >= 0 && 0 >= 1 + C && A >= 1 + D] 7. start0(A,B,C,D) -> start(A,B,B,A) [A >= 0 && B >= 0] (1,1) Signature: {(a,4);(b,4);(cont1,4);(start,4);(start0,4);(stop1,4);(stop2,4);(stop3,4)} Flow Graph: [0->{},1->{2,3},2->{},3->{4},4->{5,6},5->{0,1},6->{},7->{0,1}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: The problem is already solved. YES(?,O(n^1))